C++ Constexpr member of template type

Question

I want to create a template class with a member which is a constexpr array. The array needs, of course, a different initialization depending on the type it is but I cannot declare the array without initializing it. The problem is that I don't know the values of array until template specialization.

//A.hpp
template<typename T>
class A {
public:
    static constexpr T a[];
    constexpr A() {};
    ~A() {};
}
//B.hpp
class B: public A<int> {
public:
    constexpr B();
    ~B();
};
//B.cpp
template<>
constexpr int A<int>::a[]={1,2,3,4,5};
B::B() {}
B::~B() {}

How can I properly initialize A::a[] in B?

Why not http://coliru.stacked-crooked.com/a/44a76f94302bd9e2 ? — Marco A., Nov 12 '14 at 12:39

score 6 · Accepted Answer · edited Nov 12 '14 at 17:29

6

Every problem can be solved by adding another layer of indirection (except too much indirection):

// no a[] here.
template <typename T> struct ConstArray;

template <> 
struct ConstArray<int> {
    static constexpr int a[] = {1, 2, 3, 4, 5};

    int operator[](int idx) const { return a[idx]; }
};

template <typename T>
class A {
    static constexpr ConstArray<T> a;
};

edited Nov 12 '14 at 17:29

Casey

41,449
7
95
125

answered Nov 12 '14 at 12:24

Barry

286,269
29
621
977

Then "Every problem can be solved by adding or removing a layer of indirection" – Tim Seguine Nov 14 '14 at 10:45

C++ Constexpr member of template type

1 Answers1